[51 Nod 1584] 加权约数和

题意

求∑i=1N∑j=1Nmax(i,j)⋅σ1(ij)\large \sum_{i=1}^N\sum_{j=1}^Nmax(i,j)\cdot\sigma_1(ij)i=1∑N​j=1∑N​max(i,j)⋅σ1​(ij)

其中

1≤N≤1061\le N\le10^61≤N≤106

1≤T≤5⋅1041\le T\le5\cdot10^41≤T≤5⋅104

σ1(n)\sigma_1(n)σ1​(n)表示nnn的约数和

题目分析

令A=∑i=1n∑j=1ii⋅σ1(ij),B=∑i=1ni⋅σ1(i2)A=\sum_{i=1}^n\sum_{j=1}^ii\cdot\sigma_1(ij),B=\sum_{i=1}^ni\cdot\sigma_1(i^2)A=∑i=1n​∑j=1i​i⋅σ1​(ij),B=∑i=1n​i⋅σ1​(i2),则Ans=2A−BAns=2A-BAns=2A−B

且有

A=∑i=1ni⋅∑j=1i∑x∣i∑y∣jx⋅j/y[(x,y)=1]=∑i=1ni⋅∑j=1i∑x∣i∑y∣jx⋅j/y∑d∣(x,y)μ(d)=∑d=1nμ(d)∑d∣xx∑x∣ii∑d∣y∑y∣jijy=∑d=1nμ(d)∑x=1⌊nd⌋dx∑x∣i⌊nd⌋di∑y=1⌊nd⌋∑y∣jidjdy=∑d=1nμ(d)d2∑i=1⌊nd⌋i∑x∣ix∑y=1⌊nd⌋∑y∣jiy=∑d=1nμ(d)d2∑i=1⌊nd⌋i⋅σ1(i)∑j=1iσ1(j)=∑d=1nμ(d)d2∑i=1⌊nd⌋i⋅σ1(i)⋅Sσ1(i)\large
\begin{aligned}
A&=\sum_{i=1}^ni\cdot\sum_{j=1}^i\sum_{x|i}\sum_{y|j}x\cdot j/y[(x,y)=1]\\
&=\sum_{i=1}^ni\cdot\sum_{j=1}^i\sum_{x|i}\sum_{y|j}x\cdot j/y\sum_{d|(x,y)}\mu(d)\\
&=\sum_{d=1}^n\mu(d)\sum_{d|x}x\sum_{x|i}i\sum_{d|y}\sum_{y|j}^i\frac jy\\
&=\sum_{d=1}^n\mu(d)\sum_{x=1}^{\lfloor\frac nd\rfloor}dx\sum_{x|i}^{\lfloor\frac nd\rfloor}di\sum_{y=1}^{\lfloor\frac nd\rfloor}\sum_{y|j}^i\frac{dj}{dy}\\
&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y=1}^{\lfloor\frac nd\rfloor}\sum_{y|j}^iy\\
&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)\sum_{j=1}^{i}\sigma_1(j)\\
&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)\cdot S_{\sigma_1}(i)\\\end{aligned}A​=i=1∑n​i⋅j=1∑i​x∣i∑​y∣j∑​x⋅j/y[(x,y)=1]=i=1∑n​i⋅j=1∑i​x∣i∑​y∣j∑​x⋅j/yd∣(x,y)∑​μ(d)=d=1∑n​μ(d)d∣x∑​xx∣i∑​id∣y∑​y∣j∑i​yj​=d=1∑n​μ(d)x=1∑⌊dn​⌋​dxx∣i∑⌊dn​⌋​diy=1∑⌊dn​⌋​y∣j∑i​dydj​=d=1∑n​μ(d)d2i=1∑⌊dn​⌋​ix∣i∑​xy=1∑⌊dn​⌋​y∣j∑i​y=d=1∑n​μ(d)d2i=1∑⌊dn​⌋​i⋅σ1​(i)j=1∑i​σ1​(j)=d=1∑n​μ(d)d2i=1∑⌊dn​⌋​i⋅σ1​(i)⋅Sσ1​​(i)​

B=∑i=1ni∑x∣i∑y∣ix⋅i/y[(x,y)=1]=∑i=1ni⋅∑x∣i∑y∣ix⋅i/y∑d∣(x,y)μ(d)=∑d=1nμ(d)∑d∣ii∑d∣x∣i∑d∣y∣ix⋅iy=∑d=1nμ(d)∑i=1⌊nd⌋di∑x∣i∑y∣idx⋅didy=∑d=1nμ(d)d2∑i=1⌊nd⌋i∑x∣ix∑y∣iiy=∑d=1nμ(d)d2∑i=1⌊nd⌋i∑x∣ix∑y∣iy=∑d=1nμ(d)d2∑i=1⌊nd⌋iσ1(i)2\large
\begin{aligned}
B&=\sum_{i=1}^ni\sum_{x|i}\sum_{y|i}x\cdot i/y[(x,y)=1]\\
&=\sum_{i=1}^ni\cdot\sum_{x|i}\sum_{y|i}x\cdot i/y\sum_{d|(x,y)}\mu(d)\\
&=\sum_{d=1}^n\mu(d)\sum_{d|i}i\sum_{d|x|i}\sum_{d|y|i}x\cdot \frac iy\\
&=\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor\frac nd\rfloor}di\sum_{x|i}\sum_{y|i}dx\cdot \frac {di}{dy}\\
&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y|i}\frac {i}{y}\\
&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y|i}y\\
&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sigma_1(i)^2\\
\end{aligned}B​=i=1∑n​ix∣i∑​y∣i∑​x⋅i/y[(x,y)=1]=i=1∑n​i⋅x∣i∑​y∣i∑​x⋅i/yd∣(x,y)∑​μ(d)=d=1∑n​μ(d)d∣i∑​id∣x∣i∑​d∣y∣i∑​x⋅yi​=d=1∑n​μ(d)i=1∑⌊dn​⌋​dix∣i∑​y∣i∑​dx⋅dydi​=d=1∑n​μ(d)d2i=1∑⌊dn​⌋​ix∣i∑​xy∣i∑​yi​=d=1∑n​μ(d)d2i=1∑⌊dn​⌋​ix∣i∑​xy∣i∑​y=d=1∑n​μ(d)d2i=1∑⌊dn​⌋​iσ1​(i)2​

∴Ans=2A−B=\large
\begin{aligned}
\therefore Ans=2A-B&=
\end{aligned}∴Ans=2A−B​=​

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